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Arctan problem help

[tex]y=tan^{-1}\left(\frac{1}{ln(x)}\right)[/tex]
[tex]y'=\frac{1}{1-\left(\frac{1}{ln(x)}\right)^2}[/tex]
[tex]y'=\frac{1}{1-(lnx)^{-2}}[/tex]

is this correct?

do I have to take into account the derivative of [itex]\frac{1}{ln(x)}[/itex]?

if I do, what is the derivative of [itex]\frac{1}{ln(x)}[/itex]?
 
Last edited:

Pyrrhus

Homework Helper
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Yes, you need to take into account the derivative of [itex] \frac{1}{\ln(x)} [/itex].

You could either do the derivative of a division or do the derivative of [itex] (\ln(x))^{-1} [/itex].
 
[tex]y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}[/tex]
[tex]y'=\frac{1}{x(lnx^2)-x}}[/tex]

is that it?
 

Pyrrhus

Homework Helper
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UrbanXrisis said:
[tex]y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}[/tex]
[tex]y'=\frac{1}{x(lnx^2)-x}}[/tex]

is that it?
Little changes...

[tex]y'=\frac{1}{1+(lnx)^{-2}}*\frac{-1}{x(\ln(x))^{2}}[/tex]

Something else i noticed

[tex] (tan^{-1} (f(x)))' = \frac{f'(x)}{1+(f(x))^2} [/tex]
 
Last edited:
can that become [tex]y'=\frac{1}{-x(lnx^2)+x}}[/tex]?
 

Pyrrhus

Homework Helper
2,160
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Stop putting the square next to the x like that, use parentheses.
 
sorry, didn't notice that

[tex]y'=\frac{1}{-x(lnx)^2+x}}[/tex]

that's what I meant
 

Pyrrhus

Homework Helper
2,160
1
Sure, except for the sign problem.
 
[tex]y'=\frac{1}{-x\ln(x)^{2}-x}[/tex]
 

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